數(shù)據(jù)倉庫與數(shù)據(jù)挖掘技術(shù)復習資料-資料下載頁
2025-04-17 01:45本頁面
【正文】 1,I2,I3,I5T9I1,I2,I3解：圖：FPtree2．給定表4所示的訓練數(shù)據(jù)，數(shù)據(jù)樣本屬性age，ine，student和credit_rating描述。類標號屬性buys_puter具有兩個不同值（即{Yes,No}）。給定一個沒有類標號的數(shù)據(jù)樣本X=(age=”=30”，ine=”medium”，student=”yes”，credit_rating=”fair”)，使用樸素貝葉斯分類預測這個數(shù)據(jù)樣本的類標號。（10分）表4 數(shù)據(jù)庫訓練數(shù)據(jù)元組RIDageinestudentCredit_ratingClass:buys_puter1=30highnofairNo2=30highnoexcellentNo331…40highnofairYes440 mediumnofairYes540lowyesfairYes640lowyesexcellentNo731…40lowyesexcellentYes8=30mediumnofairNo9=30lowyesfairYes1040mediumyesfairYes11=30mediumyesexcellentYes1231…40mediumnoexcellentYes1331…40highyesfairYes1440mediumnoexcellentNo解：（1）每個類的先驗概率P(buys_puter=”Yes”)=9/14P(buys_puter =”No”)=5/14（2）為計算P(X/Ci)，i=1,2，計算下面的條件概率：P(age=”30”| buys_puter =”Yes”)=2/9P(age=”30”| buys_puter =”No”)=3/5P(ine=”medium”| buys_puter =”Yes”)=4/9P(ine=”medium”| buys_puter =”No”)=2/5P(student=”yes”| buys_puter =”Yes”)=6/9P(student=”yes”| buys_puter =”No”)=1/5P(credit_rating=”fair”| buys_puter =”Yes”)=6/9P(credit_rating=”fair”| buys_puter =”No”)=2/5（3）總結(jié)：使用以上概率，可以得到：P(X| buys_puter =”yes”)=(2/9)*(4/9)*(6/9)*(6/9)=P(X| buys_puter =”No”)=(3/5)*(2/5)*(1/5)*(2/5)=P(X| buys_puter =”yes”)P(buys_puter =”yes”)=*(9/14)=P(X| buys_puter =”No”) P(buys_puter =”No”)=*(5/14)=因此，對于樣本X，樸素貝葉斯分類預測buys_puter =”Yes”。第 10 頁 共 10 頁
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